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I have two equations:

(1) Eq1 = -((A N1 P (A B k (a N1 + aa P - r) + (d + A B r) R))/k)

(2) Eq2 = R*N1*P, where 

R = (k (aa B m - a mm + A B r))/d;
P = (-a^2 b k mm - A B m r + a k (aa bb m + A b B r))/(A d);
N1 = (-aa^2 bb k m + A mm r + aa k (a b mm - A bb r))/(A d);

I want to compare (Eq2) to (Eq1) to find the missing elements. Is there an easy way to do that in mathematica?

The answer is Eq1 = -((A N1 P (A B k (a N1 + aa P - r) + (d + A B r) R))/k) is equal to - A R P N1 d/k (i.e. Eq2 needs to be multiplied be (-Ad/k)

I tried doing FullSimplify[Eq1/Eq2] which is -((A (-a d mm + A^2 B^2 r^2 + aa B (d m + a A (b B - bb) k r)))/(k (aa B m - a mm + A B r))) that doesn't help me so much.

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  • Possible duplicate of [How to substitute functions and Full Simplify in Mathematica?](https://stackoverflow.com/questions/56466903/how-to-substitute-functions-and-full-simplify-in-mathematica) – whuber Jun 06 '19 at 20:41

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